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Polar Coordinates

- Calculus
- Absolute Maxima and Minima
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- Application of Derivatives
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Imagine that you and a friend are making your way toward a coffee shop in the middle of a large city. Looking at a map, your friend says that you are about \(1\) mile (\(1.61 km\)) southwest of the coffee shop. You take a look at the same map and see that you are \(4\) blocks across and \(4\) blocks down from the coffee shop.

In this scenario, you and your friend are both communicating the same information about where you are with respect to the coffee shop. The difference is that, while you used a **rectangular coordinate system** or **Cartesian coordinates**, your friend used a **polar coordinate system**. This article will introduce polar coordinates, how to graph polar coordinates, how to convert between polar and rectangular coordinates, and generalizations of polar coordinates in three dimensions.

You are probably used to talking about points in the plane by referring to their \(x\) and \(y\) coordinates. This system of labeling points is called the **rectangular coordinate system, **also known as **Cartesian coordinates**.

**Polar coordinates** are just a different way of labeling points in the plane. Instead of using \(x\) and \(y\) coordinates, the **polar coordinate system** labels points by how far they are from the origin, \(r\), and their angle \(\theta\) with respect to the positive \(x\)-axis, measured counterclockwise. This angle \(\theta\) is typically measured in radians.

Further below in this article, you can learn how to convert between degrees and radians, just in case you forgot!

In principle, both the angle \(\theta\) and the **radial distance** \(r\) can take on any real value. In practice, \(r\) values are expressed using only non-negative values, that is

\[ r \in [0,\infty).\]

If \(r\) is zero, then, regardless of \(\theta\), the point \((r,\theta)\) is at the pole.

The usual convention is to express it as if \(\theta=0\) to avoid any ambiguity.

As mentioned before, \(r\) can take on negative values, but this is usually not expressed like this. Instead of writing a point \((-r,\theta)\) with a negative \(r\) coordinate, you write the reflection of the point \((r,\theta)\) by adding \(\pi\) to the angle.

The angular coordinate, \(\theta\), is typically expressed using values between \(0\) and \(2\pi\), that is

\[ \theta \in [0,2\pi).\]

Just like the radial coordinate, it is also possible to get a point \((r,-\theta)\) with a negative \(\theta\) coordinate. You can obtain this point by rotating clockwise from the positive x-axis by \(\theta\) instead of counterclockwise.

Polar and Cartesian coordinates implicitly introduce the notion of different **metrics** and different ways of talking about distance. Going back to our coffee shop scenario, you could either say that you and your friend are \(8\) blocks away from the coffee shop (4 blocks across and \(4\) blocks down), or you are \(1\) mile (\(1.61 km\)) away from the coffee shop. Both notions of distance have their advantages. The 'block distance,' called the **taxicab metric**, tells you how much you need to walk to get to the coffee shop if you must stay on the sidewalk. The 'diagonal distance,' or **Euclidean metric**, tells you the shortest path you could possibly take to get to the coffee shop, assuming you can cut diagonally across city blocks.

Metrics and their associated **metric spaces** become increasingly important as you continue to study math and have incredible practical applications in areas ranging from data science to signal processing to quantum mechanics.

There are several terms that are important to know when working with polar coordinates. The point we call the **origin** when working with rectangular coordinates is called the **pole** when working with polar coordinates. What we call the x-axis when working with rectangular coordinates is called the **polar axis** when working with polar coordinates.

The polar axis is also known as the **reference direction**.

The distance from the origin denoted as the \(r\) coordinate, is also referred to as the **radius**, **radial coordinate**, or **radial distance**. The \(\theta\) coordinate is called the **angular coordinate**, **polar angle**, or **azimuth**.

Table 1. Terminology used in polar coordinates

Name(s) in Rectangular Coordinates | Name(s) in Polar Coordinates |

Origin | Pole |

Horizontal axis, or \(x\)-axis | Polar axis, or reference direction |

Distance from the origin | Radial coordinate, radius, or radial distance |

Counterclockwise angle from the positive \(x\)-axis | Angular coordinate, polar angle, or azimuth |

The angular coordinate \(\theta\) is generally specified in either degrees or radians. Degrees and radians are just different ways of measuring angles.

Algebraically, the relationship between degrees and radians is as follows:

\[\begin{align} k\pi\text{ radians } &= 180k \text{ degrees } \\ k\text{ degrees } &= \dfrac{k\pi}{180} \text{ radians }\end{align}\]

The angle \(30^{\circ}\) can be written in terms of radians. Using the conversion

\[k\text{ degrees } = \dfrac{k\pi}{180}\text{ radians}, \]

you will get that

\[ \begin{align} 30\text{ degrees } &= \dfrac{30\pi}{180}\text{ radians } \\ &= \dfrac{\pi}{6}\text{ radians}. \end{align}\]

Conversely, an angle of \(\tfrac{\pi}{4}\) radians can be written as degrees. Using the conversion

\[k\pi\text{ radians } = 180k \text{ degrees}, \]

you will get that

\[\begin{align} \frac{\pi}{4}\text{ radians } &= \frac{1 \, \pi \text{ radians}}{4} \\ &= \dfrac{180\text{ degrees}}{4} \\ &= 45 \text{ degrees}. \end{align}\]

One of the interesting aspects of working with polar coordinates is that every point has infinitely many polar coordinates that describe it. You noted this earlier for the pole, which we said is represented by \((0,\theta)\) for any value of \(\theta\). For example, the coordinates \((0,0)\), \((0,-\pi)\), and \(\left(0,\tfrac{\sqrt{\pi}}{17}\right)\) all represent the pole.

It is worth noting that this multivalued representation is usually avoided and the coordinates are shown in the range

\[ r \in [0,\infty)\]

and

\[ \theta \in [0,2\pi).\]

However, for some special scenarios, like the description of motion in Physics, these domains are extended so both include all real numbers.

In general, the points \((r,\theta)\) and \((r,\theta+2n\pi)\) for any integer \(n\) describe the same point. The points \((-r,\theta)\) and \((r,\theta+m\pi)\), where \(m\) is an *odd *integer, also describe the same point.

Geometrically, \((r,\theta)\) and \((r,\theta+2n\pi)\) represent the same point because adding \(2n\pi\) corresponds to rotating the point by \(2n\pi\) radians, or some multiple of 360 degrees. Rotating by a multiple of \(2\pi\) effectively does not change the position of the point, so \((r,\theta)\) and \((r,\theta+2n\pi)\) represent the same point.

The coordinates \(\left(1,\tfrac{\pi}{4}\right)\) and \(\left(1,\tfrac{9\pi}{4}\right)\) describe the same point since

\[ \begin{align} \frac{9\pi}{4} &=\frac{\pi}{4}+\frac{8\pi}{4}\\ &=\frac{\pi}{4}+2\pi. \end{align}\]

Likewise, the coordinates \(\left(-10,\tfrac{5\pi}{6}\right)\) and \(\left(10,\tfrac{11\pi}{6}\right)\) describe the same point, since

\[\begin{align} \frac{11\pi}{6} &= \frac{5\pi}{6}+\frac{6\pi}{6} \\ &=\frac{5\pi}{6}+\pi. \end{align}\]

By this point, you may already notice that working with polar coordinates has its advantages, so here you can learn how to convert from rectangular coordinates to polar coordinates.

Given a point in Cartesian coordinates, you can find an expression for that same point in polar coordinates. To do so, you can use the formulas

\[\begin{align}r&=\sqrt{x^2+y^2} \\ \theta &= \arctan{\left( \frac{y}{x}\right)}. \end{align} \]

You can also find the arctangent function written as the inverse tangent function, that is

\[ \theta = \tan^{-1}{\left( \frac{y}{x}\right)}.\]

To see where these formulas come from, say you are given a point \((x,y)\) in rectangular coordinates and you want to know its corresponding polar coordinates \((r,\theta)\). If you were to graph the point, you would get something like the following image.

You can use known facts about right triangles to write the relationships between \(x\), \(y\), \(r\), and \(\theta\). First, notice that you can use the Pythagorean Theorem to write

\[x^2+y^2=r^2.\]

From here, you can solve for \(r\) to obtain the first formula,

\[ r = \sqrt{x^2+y^2}.\]

You can also use the fact that the tangent of an angle \(\theta\) in a right triangle is the ratio of its opposite side to its adjacent side so that you can write

\[ \tan{\theta} = \frac{y}{x}.\]

You can now use the inverse tangent function to isolate \(\theta\), that is

\[ \theta = \arctan{\left( \frac{y}{x}\right)}.\]

The point

\[ P=(3, 4)\]

is given in cartesian coordinates. Write \(P\) in polar coordinates.

**Solution:**

To find \(r\) you just need to square each component of the point, add them together, and take the square root, that is

\[ \begin{align} r &= \sqrt{x^2+y^2} \\ &= \sqrt{3^2+4^2} \\ &= \sqrt{ 9 +16} \\ &= \sqrt{25} \\ &= 5. \end{align}.\]

Next, find the angle by using

\[\theta = \arctan{\left(\frac{y}{x}\right)}\]

with the help of a calculator, that is

\[ \begin{align} \theta &= \arctan{\frac{y}{x}} \\ &= \arctan{\left(\frac{4}{3} \right)} \\ &= 0.9272. \end{align} \]

Remember that this angle is typically given in radians, so make sure your calculator uses these!

This means that the point \(P\) in polar coordinates is

\[ P =(5,0.9272).\]

Since you need to use the inverse tangent function to find the angle \(\theta\), it is better to discuss this a bit further.

When using the formula

\[\theta=\arctan{\left(\frac{y}{x}\right)},\]

you have to be aware of certain technicalities related to the tangent and arctangent functions.

First, notice that this expression is undefined when \(x=0\). In this case, the point \((x,y)\) must lie on the \(y\)-axis. So, \(\theta\) must either be \(\tfrac{\pi}{2}\) or \(-\tfrac{\pi}{2}\), depending on whether \(y\) is positive or negative.

Next, given any \(\tfrac{y}{x}\), there are actually two values of \(\theta\) in \([-\pi,\pi]\) satisfying \(\tan(\theta)=\tfrac{y}{x}\).

Given \(y=1\) and \(x=1\), both the angles \(\tfrac{\pi}{4}\) and \(-\tfrac{3\pi}{4}\) satisfy that

\[ \begin{align} \tan\left(\dfrac{\pi}{4}\right) &=\tan\left(-\dfrac{3\pi}{4}\right) \\ &=\dfrac{y}{x} \\ &=\dfrac{1}{1} \\&=1. \end{align}\]

Notice that these angles are reflections of each other; this is always the case.

To make sure that you are specifying the correct point when writing a point in polar coordinates, you must make sure that the angle \(\theta\) you use is in the correct quadrant. The arctangent function only returns angles between \(-\tfrac{\pi}{2}\) and \(\tfrac{\pi}{2}\), so \(\arctan\left(\tfrac{y}{x}\right)\) only returns the correct value for \(\theta\) if the point \((x,y)\) is in the first or fourth quadrant.

Furthermore, since \(\theta\) values are typically given between \(0\) and \(2\pi\), you also need to do a correction if the angle you obtain from the calculator is negative. The following table summarizes how to find \(\theta\).

Table 2. Relation between the sign of \(x\) and the angle \(\theta\)

Quadrant | Sign of \(x\) | Sign of \(y\) | \( \theta\) |

I | \(+\) | \(+\) | \(\theta=\arctan\left(\dfrac{y}{x}\right)\) |

II | \(-\) | \(+\) | \(\theta=\arctan\left(\dfrac{y}{x}\right)+\pi\) |

III | \(-\) | \(-\) | \(\theta=\arctan\left(\dfrac{y}{x}\right)+\pi\) |

IV | \(+\) | \(-\) | \(\theta=\arctan\left(\dfrac{y}{x}\right)+2\pi\) |

Consider the point \( P= (-2,1) \).

- In which quadrant is \(P\)?
- Find its \(r\) coordinate.
- Find its \(\theta\) coordinate.

**Solution:**

- Since the \(x\)-coordinate is negative and its \(y\)-coordinate is positive, the point \(P\) lies in the
**second quadrant (II)**. - This step is rather straightforward, just use the formula for \(r\), that is\[ \begin{align} r &= \sqrt{x^2+y^2} \\ &= \sqrt{(-2)^2+1^2} \\&= \sqrt{4+1} \\&= \sqrt{5}.\end{align} \]
- You just found that \(P\) is on the second quadrant, so use a calculator and find that\[\begin{align} \theta &= \arctan{\left( \frac{1}{-2}\right)}+\pi \\ &= \arctan{\left(-0.5\right)}+\pi \\ &= -0.463647+3.141592 \\ &= 2.677945.\end{align}\]

Now suppose you are given a point in polar coordinates and you want to know how to express this point using rectangular coordinates. For this case, you can also use the same diagram of the right triangle that relates \(x\), \(y\), \(r\), and \(\theta\).

Focus on the angle \(\theta\). Since you have a right triangle, you can write the sine of the angle as:

\[ \sin{\theta} = \frac{y}{r}\]

Likewise, the cosine of the angle is:

\[ \cos{\theta} = \frac{x}{r}\]

Solving each expression for \(x\) and \(y\) will yield you:

\[ x=r\,\cos{\theta}\]

and

\[ y=r\,\sin{\theta}\]

Unlike when passing from rectangular coordinates to polar coordinates, the above expressions do not need you to watch out for any special considerations. These are as straightforward as it gets!

You previously found that the point \(P=(-2,1) \) had:

\[ r = \sqrt{5}\]

and

\[ \theta = 2.677945 \]

Verify that the above values are correct.

**Solution:**

To find the \(x\)-coordinate use:

\[ x = r \, \cos{\theta},\]

so use a calculator and find:

\[ \begin{align} x &= \sqrt{5} \cdot \cos{2.677945} \\ &= -2. \end{align}\]

For the \(y\)-coordinate, use instead:\[ y = r\, \sin{\theta}\]

that is

\[ \begin{align} y &= \sqrt{5} \cdot \sin{2.677945} \\ &= 1. \end{align}\]

The above values are precisely the rectangular coordinates of the point \(P\).

You can use several electronic resources to graph points in polar coordinates. For example, in Geogebra, you can graph a point \((r,\theta)\) by typing \((r;\theta)\). Many graphing calculators also allow you to plot points using polar coordinates, as do programs and programming languages like Python, Octave, and Matlab. For information on how to graph polar curves, see the article Polar Curves.

Here you can look at some examples of conversion between both coordinate systems!

For these examples, suppose you are given a point in rectangular coordinates and you want to find it written in polar coordinates.

Convert the point \[ P= \left(-2\sqrt{3},-2\right)\]

from rectangular to polar coordinates.

**Solution:**

First, note that this point is in the third quadrant since both \(x\) and \(y\) are negative.

Next, find \(r\) using

\[r=\sqrt{x^2+y^2},\]

that is

\[\begin{align}r&=\sqrt{\left(-2\sqrt{3}\right)^2+(-2)^2} \\ &=\sqrt{4(3)+4} \\ &= \sqrt{12+4} \\ &= \sqrt{16}\\ &=4.\end{align}\]

Next, find theta. Since the point is on the third quadrant you will need to use\[\theta=\arctan\left(\frac{y}{x}\right)+\pi,\]

that is

\[\begin{align}\theta&=\arctan\left(\frac{-2}{-2\sqrt{3}}\right)+\pi \\ &=\arctan\left(\dfrac{1}{\sqrt{3}}\right)+\pi \\ &=\dfrac{\pi}{6}+\pi \\ &=\dfrac{7\pi}{6}.\end{align}\]

Thus, the point \(\left(-2\sqrt{3},-2\right)\) in polar coordinates is \(\left(4,\tfrac{7\pi}{6}\right)\).

Here is another example.

Convert the point

\[ Q = \left(\tfrac{3\sqrt{2}}{2},-\tfrac{3\sqrt{2}}{2}\right)\]

from rectangular to polar coordinates.

**Solution:**

Begin by noting that this point is in the fourth quadrant because \(x\) is positive and \(y\) is negative.

Now find \(r\) as usual, that is

\[\begin{align} r &= \sqrt{\left(\dfrac{3\sqrt{2}}{2}\right)^2+\left(-\dfrac{3\sqrt{2}}{2}\right)^2} \\ &= \sqrt{\dfrac{9\cdot2}{4}+\dfrac{9\cdot2}{4}} \\ &=\sqrt{\dfrac{9}{2}+\dfrac{9}{2}}\\ &=\sqrt{9} \\ &=3.\end{align}\]

You previously found that the point is on the fourth quadrant, so you will need to use

\[ \theta = \arctan{\left( \frac{y}{x} \right)}+2\pi.\]

Before proceeding, note that the \(x\) and \(y\) values are the same, just with opposite signs. This means that

\[ \frac{y}{x}=-1,\]

so you do not need to write the whole huge fraction, as it will simplify to \(-1\). Knowing this,

\[ \begin{align} \theta &= \arctan{-1}+2\pi \\ &= -\frac{\pi}{4}+2\pi \\ &= \frac{7\pi}{4}.\end{align}\]

This means that the point \(Q\) written in polar coordinates is

\[ Q = \left(3, \frac{7\pi}{4}\right).\]

For the next example, suppose you are given a point in polar coordinates and you wish to know how it's written in rectangular coordinates.

Convert the point

\[\left(4,-\tfrac{\pi}{3}\right)\]

from polar to rectangular coordinates.

**Solution:**

Start by finding \(x\) using the formula \(x=r\cos(\theta)\).

\[\begin{align} x&=4\cos\left(-\dfrac{\pi}{3}\right) \\ &= 4\left(\dfrac{1}{2}\right) \\ &=2.\end{align}\]

Next, find \(y\), using the formula \(y=r\sin(\theta)\).

\[\begin{align} y&=4\sin\left(-\dfrac{\pi}{3}\right) \\ &= 4\left(-\dfrac{\sqrt{3}}{2}\right) \\ &=-2\sqrt{3}.\end{align}\]

So, the point \(\left(4,-\tfrac{\pi}{3}\right)\) in Cartesian coordinates is \((2,-2\sqrt{3})\).

Here comes one last example.

Convert the point

\[\left(5,\tfrac{3\pi}{4}\right)\]

from polar to rectangular coordinates.

**Solution:**

First, find \(x\) using the formula \(x=r\cos(\theta)\).

\[\begin{align} x&=5\cos\left(\dfrac{3\pi}{4}\right) \\ &= 5\left(-\dfrac{\sqrt{2}}{2}\right) \\ &=-\dfrac{5\sqrt{2}}{2}.\end{align}\]

Then, find \(y\), using the formula \(y=r\sin(\theta)\).

\[\begin{align} x&=5\sin\left(\dfrac{3\pi}{4}\right) \\ &= 5\left(\dfrac{\sqrt{2}}{2}\right) \\ &=\dfrac{5\sqrt{2}}{2}.\end{align}\]

Therefore, the point \(\left(5,\tfrac{3\pi}{4}\right)\) in Cartesian coordinates is \(\left(-\tfrac{5\sqrt{2}}{2},\tfrac{5\sqrt{2}}{2}\right)\).

Don't worry if your calculator is not giving you the exact values of the trigonometric functions. If you give a decimal value the answer is still the same.

- The polar coordinate system is a system of labeling points in the plane by their distance from the origin and their angle with respect to the positive x-axis.
- There are multiple ways to represent any given point in polar coordinates.
- When converting from rectangular to polar coordinates, be careful about which quadrant your point is in, and be sure to use the appropriate arctangent function.

You need only to apply the relations x=rcosθ and y=rsinθ.

More about Polar Coordinates

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